Problem 63 - A Graphical Approach to College Algebra

Question

Solve each problem. Planetary Orbits The formula $$ f(x)=x^{1.5} $$ calculates the number of years it would take for a planet to orbit the sun if its average distance from the sun is \(x\) times farther than Earth. If there were a planet located 15 times farther from the sun than Earth, how many years would it take for the planet to orbit the sun?

Step-by-Step Solution

Answer

It takes approximately 58.1 years for the planet to orbit the sun.

Step 1: Identify the Given Formula

We are given the formula \( f(x)=x^{1.5} \), which can calculate the orbital period of a planet, in years, based on its average distance from the sun relative to Earth.

Step 2: Substitute the Given Value

The problem states that the planet is 15 times farther from the sun than Earth, so we substitute \( x = 15 \) into the formula. The expression becomes \( f(15) = 15^{1.5} \).

Step 3: Calculate the Exponent

To find \( 15^{1.5} \), understand that \( 1.5 \) is the same as \( 3/2 \), so we can rewrite this as \( 15^{3/2} \). This breaks down into \( \sqrt{15^3} \).

Step 4: Calculate Inside the Radical

Calculate \( 15^3 \) first. \( 15 \times 15 = 225 \) and \( 225 \times 15 = 3375 \). Thus, \( 15^3 = 3375 \).

Step 5: Calculate the Square Root

Next, find \( \sqrt{3375} \). By approximation, this value is around 58.0947. Therefore, \( f(15) = 58.0947 \).

Step 6: Conclude the Calculation

Based on the calculation, it would take approximately 58.1 years for the planet 15 times farther from the sun than Earth to complete one orbit.

Key Concepts

Exponents

Understanding exponents is crucial in many mathematical applications, especially when dealing with repeated multiplication. An exponent of a number shows how many times the number, known as the base, is multiplied by itself. For instance, in the expression \( x^{1.5} \), the base is \( x \) and the exponent is 1.5, indicating that \( x \) is multiplied by itself one and a half times.

It’s helpful to remember the basic rules of exponents:

• \( a^1 = a \): Any number raised to the power of 1 is itself.
• \( a^0 = 1 \): Any number raised to the power of 0 is 1.
• \( a^{m+n} = a^m \times a^n \): Adding exponents entails multiplying the bases.

In situations like our exercise, we often encounter non-integer exponents such as 1.5, which can also be expressed as \( \frac{3}{2} \). This is an important step in dealing with these expressions, allowing us to use both multiplication and radical operations.

Radicals

Radicals arise when we need to find the root of a number, as seen in the conversion from \( 15^{1.5} \) to \( \sqrt{15^3} \). A square root symbol, \( \sqrt{} \), indicates we are looking for a number which, when multiplied by itself, yields the number inside the radical.

Here's what you need to know about radicals:

• A radical can be represented as a fractional exponent. For example, \( \sqrt{x} \) is equivalent to \( x^{0.5} \).
• The product inside the radical should first be calculated, as we did with \( 15^3 \) turning into 3375.
• Finding the square root can be complex without a calculator, but approximation helps in many cases to simplify matters.

In our exercise, we computed \( \sqrt{3375} \) to transform \( 15^{1.5} \) into an approximate value of 58.0947, signifying the number of years for a planetary orbit.

Orbital Period Calculation

Orbital period calculation is an interesting application of mathematical principles that helps us understand celestial mechanics. The formula \( f(x) = x^{1.5} \) gives the time it takes for a planet to orbit the sun based on the distance compared to Earth's orbit.

Here are some key concepts:

• The formula uses the fact that the time for an orbit is proportional to the distance raised to the power of 1.5. This is derived from Kepler's Third Law in celestial mechanics.
• Substituting specific values, like \( x = 15 \), helps us practically apply this formula to understand how it works.
• The implication of this calculation is significant in astronomy, allowing us to predict orbits of planets, asteroids, and other celestial bodies.

In the given problem, after computing \( 15^{1.5} \), we found that such a planet would take about 58.1 years to complete its revolution around the sun. This emphasizes how distance and time are interlinked in planetary orbits.